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STRUCTURE OF Pd-102, Pd-104, Pd-105, Pd-106, Pd-108, Pd-110
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPT . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Structure of Palladium (Pd) with 22 blank positions Naturally occurring palladium (Pd) is composed of six stable isotopes, Pd-102, Pd-104, Pd-105, Pd106, Pd108, and Pd-110, although two of them are theoretically unstable. The most stable radioisotopes are Pd-107 with a half-life of 6.5 million years, Pd-103 with a half-life of 17 days, and Pd-100 with a half-life of 3.63 days. In general Palladium-92 with 46 protons and 46 neutrons (even number of protons and neutrons ) has the same structure of Ru characterized by the high symmetry. (See my STRUCTURE OF Ru-96....Ru-104 ). Since the shape of Ru with two squares is elongated the additional p45n45 and p46n46 as two pn systems with vertical strong bonds make at symmetrical positions two vertical rectangles outside the central parallelepiped formed by p41n41, p42n42, p43n43 and p44n44. Of course the vertical bonds occur with S=0. In the following diagram of Pd-92 you see that the additional p45n45 and p46n46 make two symmetrical rectangles with n25 and p27 and p26 and n28. Note that the two protons of the symmetrical vertical rectangles like the p45 and p46 make at the parallelepiped ( formed by the third and the fourth horizontal plane) 4 blank positions for receiving 4 extra (n) with weak bonds. Also the p45 and p46 contribute for the creation of 2 more blank positions able to receive 2n with strong bonds at the second and the fifth horizontal plane. For example looking at the top view of the third horizontal plane one concludes that the proton p45 with p25 and p41 makes two blank positions for receiving extra 2(n) with weak horizontal bonds. Also the p45 and p46 combined with p23 and p30 respectively make the blank positions to receive two extra n at the second and the fifth horizontal plane. At the same horizontal planes one observes also 4 blank positions able to receive 4for making three bonds per neutron. In this arrangement one observes also that the shape has the two deuterons like p38n38 and p40n40 with S =+2 and the two deuterons like p37n37 and p39n39 with S = -2. Note that the protons of the two squares form 8 blank positions able to receive 8 [n with strong bonds. Then the protons of the first and the sixth horizontal plane form 4 blank positions able to receive 4 (n) with weak bonds. That is, the total number N of blank positions able to receive N neutrons is given by The two squares which can receive 8n The central parallelepiped which can receive 4(n) The second and the fifth H. plane which can receive 4{n} + 2n The first and the sixth H. Plane which can receive 4(n) Thus N = 8n + 4(n) + 4{n} + 2n + 4(n) = 22 ' ' ' ' STRUCTURE OF Pd-102, Pd-104, Pd-106, Pd-108, AND Pd-110 WITH S = 0 ' '''The Pd-102 of S=0 has 10 extra neutrons. Therefore it makes the stable structure by receiving 4{n} and 6n with opposite spins. Also the Pd-104 which has 12 extra neutrons receives 4{n} and 8n of opposite spins. In the same way the Pd-106 the Pd-108 and the Pd-110 receive 14, 16, and 18 extra neutrons respectively for making stable structures. '''STRUCTURE OF Pd-105 WITH S =+5/2' For symmetrical arrangements in this case in the Pd-92 the p39n39 changes the spin from S=-1 to S= +1 in order to fill the symmetrical blank position behind the p40n40. This change gives S = +2. which differs from the well-known Pd-92 with S =0. Under this condition since Pd-105 has 13 extra neutrons one concludes that the total spin S = +5/2 is due to seven extra neutrons of positive spins and to six extra neutrons of negative spins. That is S = +2 + 7(+1/2) + 6(-1/2) = +5/2 ' ' ' DIAGRAM OF Pd-92 WITH 22 BLANK POSITIONS' Here you see the additional n45p45 and the n46p46 of opposite spins which make symmetrical rectangles. Whereas the p41, n41, p42, n42, p43, n43 p44, and n44 of opposite spins which make the central great parallelepiped are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Note that the 4 extra neutrons (n) of the first and the sixth plane along with the 4 extra neutrons n near the p37, p38, p39 and p40 are not shown. You can see only the 4 extra neutrons n existing under the p21 and p22 and over the p31 and p32 ' ' n40.........p40.....n ' ' n........p38..........n38 H. square with n ' ' n31………p12.........n12.......p32 ' p31........n11.........p11…… n32 Sixth H. plane' ' p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30 Fifth H. plane' ' n27.........p8..........n8.........p28' ' n45.....p27.........n7..........p7........n28...........p46 Fourth H. plane' ' p25.........n6.........p6..........n26' ' p45....n25……….p5........n5……….p26.........n46 Third H. plane' ' n23………p4........n4…….p24' ' p23……..n3………p3……….n24 Second H. plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n' ........p37.. ....n37 ' ' n39.......p39........n ' H. square with n' ' ' ' ' ' TOP VIEW OF THE SQUARE OF p37n37 AND n39p39 ' THE TWO EXTRA NEUTRONS n WITH STRONG VERTICAL BONDS UNDER THE p21 AND p22 ARE ALSO SHOWN IN THE DIAGRAM , WHILE THE TWO EXTRA NEUTRONS nWITH STRONG VERTICAL BONDS UNDER THE p34 AND p33 ARE NOT SHOWN IN THE DIAGRAM ' n' ' n........P37.........n37 ' ' n39........p39.........n ' ' n' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' ' ' ' ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' TOP VIEW OF THE THIRD HORIZONTAL PLANE OF POSITIVE SPINS WITH THE 4 NUCLEONS LIKE p41, n43, n42 AND p44 WHICH MAKE THE SQUARE . HERE THE p45 AND n46 ALONG WITH n45 AND p46 OF THE FOURTH PLANE MAKE THE SYMMETRICAL RECTANGLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n15p15 AND p16n16. ' ' n42........p16......n16......p44 ' p25........n6........p6........n26' ' p45........n25........p5........n5........p26........ ' Category:Fundamental physics concepts